Question

Q.A stone, of mass m, is attached to a strong string and whirled in a vertical circle of radius r. At the exact bottom of the path the tension in the string is 3 times the stone's weight. The stone's speed at this point is given by ...?
A. 2gr)1/2.

please explain everything in detail

Answer 1

Q.A stone, of mass m, is attached to a strong string and whirled in a vertical circle of radius r. At the exact bottom of the path the tension in the string is 3 times the stone's weight. The stone's speed at this point is given by ...?

A. $2\mathrm{gr})1 / 2$

please explain everything in detail

Solution:

Newton's second law for the stone At the exact bottom of the path:

$m\vec{g} + \vec{T} = m\vec{a}_c = \vec{F}_{centr}$

$mg$ - stone's weight;

$T = 3mg$ - tension in the string;

$F_{centr}$ - Centripetal force (the stone is going in a circle);

$a_{c}$ - centripetal acceleration.

Projection on the Y-axis:

$$y: T - m g = m a _ {c}$$

at the bottom the tension is is 3 times the stone's

weight:

$$T = 3 m g$$

Formula for centripetal acceleration:

$$a _ {c} = \frac {V ^ {2}}{r}$$

(3)and(2)in(1):

$$3 m g - m g = m \frac {V ^ {2}}{r}$$ $$2 m g = m \frac {V ^ {2}}{r}$$ $$2 g r = V ^ {2}$$ $$V = \sqrt {2 g r}$$

Answer: The stone's speed at the exact bottom of the path is given by $V = \sqrt{2gr}$